Question

What happens to the \(K_{\mathrm{sp}}\) value of a solid as the temperature of the solution changes? Consider both increasing and decreasing temperatures, and explain your answer.

Short answer

The effect of temperature on the $K_{\mathrm{sp}}$ value of a solid depends on whether the dissolution process is endothermic or exothermic. For endothermic processes, $K_{\mathrm{sp}}$ increases with increasing temperature and decreases with decreasing temperature. For exothermic processes, $K_{\mathrm{sp}}$ decreases with increasing temperature and increases with decreasing temperature. This can be understood using Le Chatelier's principle and the relationship between Gibbs free energy change and the equilibrium constant.

Step-by-step solution

Understanding the Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is a measure of how much of a slightly soluble ionic compound can dissolve in water at a particular temperature. It is defined as the product of the concentrations of the ions in the solution, each raised to the power of their stoichiometric coefficients in the balanced chemical equation. For example, for the following solubility equilibrium of a generic compound AB: \(AB (s) \rightleftharpoons A^{+} + B^{-}\) The solubility product constant expression would be: \(K_{\mathrm{sp}} = [A^{+}][B^{-}]\)

Considering the Effect of Temperature on Ksp

To determine how temperature affects the solubility product constant, we will consider the relationship between Ksp and the changes in temperature using Le Chatelier's principle and Gibbs free energy. Le Chatelier's principle states that if a system at equilibrium is subjected to a change in temperature, pressure, or concentration of components, the system will adjust to counteract the change and reestablish a new equilibrium. In the context of solubility, temperature is the factor we are interested in. Gibbs free energy change, ΔG, is related to the equilibrium constant K (in our case, Ksp) by the equation: \(\Delta G = -RT \ln{K}\) where R is the universal gas constant, T is the temperature in Kelvin, and K is the equilibrium constant (Ksp for solubility).

Analyzing the Effects of Increasing and Decreasing Temperature

When the temperature increases, the system tries to counteract the change by shifting the equilibrium in the direction that absorbs heat. If the dissolution of the solid is an endothermic process (entropy change, ΔH, is positive), then more of the solid will dissolve in water, increasing the Ksp value. On the other hand, if the dissolution is an exothermic process (ΔH is negative), the solubility will decrease with increasing temperature, and Ksp will decrease. Conversely, when the temperature decreases, the system tries to counteract the change by shifting the equilibrium in the direction that releases heat. If the dissolution of the solid is an endothermic process, less of the solid will dissolve in water, decreasing the Ksp. If the dissolution is exothermic, the solubility will increase with decreasing temperature, and Ksp will increase. In summary, the effect of temperature on the Ksp value of a solid depends on whether the dissolution process is endothermic or exothermic. For endothermic processes, Ksp increases with increasing temperature and decreases with decreasing temperature. For exothermic processes, Ksp decreases with increasing temperature and increases with decreasing temperature.